How do you solve $4y - ((3y+4)/2) + 1/3 = 5((4-y)/3)$ ?

Question

1925 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-24T10:03:05

$color(red)(y = 2$

$4y - ((3y + 4)/2) + 1/3 = 5 ((4-y)/3)$

$4y - (3y)/2 -2 + 1/3 = 20/3 - (5y)/3, " removing braces"$

$4y - (3y)/2 + (5y) / 3 = 20/3 + 2 - 1/3, "rearranging like terms together"$

$(24y - 9y + 10y) / 6 = (20 + 6 - 1) / 3, " taking L C M"$

$(25 y ) / cancel(6)^color(red)(2) = 25 / cancel3, " simplifying"$

$25y = 25 * 2 = 50, " cross multiplying"$

$color(red)(y = 50 / 25 = 2$

How do you solve 4y - ((3y+4)/2) + 1/3 = 5((4-y)/3)4y - ((3y+4)/2) + 1/3 = 5((4-y)/3)?